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2 years ago

For the common-emitter, common-base and emitter-follower amplifier designs,what is the primary benefit of each amplifier?

Physics

1 answer:

kati45 [8]2 years ago

6 0

Answer:

Explanation:

A common emitter amplifier works by inverting. It does have low input impedance. Despite its low input impedance, it poses an otherwise high output impedance.

The common base circuit performs optimally when it acts as a current buffer. It has the ability to take an input current at a low input impedance, and transmit almost the same current to an impedance with a higher output

The primary benefit of emitter follower amplifier is that the transistor is able to provide current and power gain. Although this transistor takes in little current from the input. It still provides an impedance with a low output to a circuit by exercising the output of the follower. This then translates to that the output under load not dropping.

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A 60kg bicyclist (including the bicycle) is pedaling to the

Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

Let's count the forces acting on the bicylist:

1) Weight ( $W=mg$ ): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force ( $F_A$ ): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag ( $R$ ): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

$F_{net}=ma$

where

$F_{net}$ is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

$a=3.1 m/s^2$ is its acceleration

Substituting, we find the net force on the bicyclist:

$F_{net}=(60)(3.1)=186 N$

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

$F_{net}=F_a-R$

where:

$F_{net}$ is the net force

$F_a$ is the applied force (forward)

$R$ is the air drag (backward)

In this problem we have:

$F_{net}=186 N$ is the net force (found in part b)

$R=60 N$ is the magnitude of the air drag

Solving for $F_a$ , we find the force produced by the bicyclist while pedaling:

$F_a=F_{net}+R=186+60=246 N$

3 0

3 years ago

Beginning<br> lonic compounds are made up of a__<br> and a__

Romashka-Z-Leto [24]

Answer:

metal and non-metal

Explanation:

Ionic bonds are formed when a metal loses electrons and a non-metal gains said electrons.

3 0

3 years ago

Read 2 more answers

the maximum intensity levels of a trumpet, trombone, and a bass drum, each at a distance of 3m are 94 dB, 107dB, and 113dB respe

Gwar [14]

Answer:

β = 114 db

Explanation:

The intensity of sound in decibles is

β = 10 log $\frac{I}{I_{o}}$

in most cases Io is the hearing threshold 1 10-12 W / cm²

let's calculate the intensity of each instrument

I / I₀ = 10 (β / 10)

I = I₀ 10 (β / 10)

trumpet

I1 = 1 10⁻¹² 10 (94/10)

I1 = 2.51 10⁻³ / cm²

Thrombus

I2 = 1 10⁻¹² 10 (107/10)

I2 = 5.01 10-2 W / cm²

low

I3 =1 1-12 (113/10) W/cm²

I3 = 1,995 10-1 W / cm²

when we place the three instruments together their sounds reinforce

I_total = I₁ + I₂ + I₃

I_ttoal = 2.51 10-3 + 5.01 10-2 + 1.995 10-1

I_total = 0.00251 + 0.0501 + 0.1995

I_total = 0.25211 W / cm²

let's bring this amount to the SI system

β = 10 log (0.25211 / 1 10⁻¹²)

β = 114 db

7 0

3 years ago

A girl throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s, and it bounc

erma4kov [3.2K]

Answer:

F = 800N

the magnitude of the average force exerted on the wall by the ball is 800N

Explanation:

Applying the impulse-momentum equation;

Impulse = change in momentum

Ft = m∆v

F = (m∆v)/t

Where;

F = force

t = time

m = mass

∆v = v2 - v1 = change in velocity

Given;

m = 0.80 kg

t = 0.050 s

The ball strikes the wall horizontally with a speed of 25 m/s, and it bounces back with this same speed.

v2 = 25 m/s

v1 = -25 m/s

∆v = v2 - v1 = 25 - (-25) m/s = 25 +25 = 50 m/s

Substituting the values;

F = (m∆v)/t

F = (0.80×50)/0.05

F = 800N

the magnitude of the average force exerted on the wall by the ball is 800N

4 0

3 years ago

Consider the following. (Let C1 = 20.80 µF and C2 = 14.80 µF.) A rectangular circuit contains a battery and four capacitors. The

umka21 [38]

Find the below attachment

7 0

3 years ago